Problem: Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

* Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.



When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

* N is an even number.
* 2 ≤ N ≤ 2 × 10^5
* p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the lexicographically smallest permutation, with spaces in between.

Examples

Input

4
3 2 4 1


Output

3 1 2 4


Input

2
1 2


Output

1 2


Input

8
4 6 3 2 8 5 7 1


Output

3 1 2 7 4 6 8 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

const int N = 200005;

int n, nn, p[N], ip[N], tot, ans[N];

struct st {
	int a[17][N / 2];
	void init() {
		for (int i = 1; i < 17; ++i)
			for (int j = 0; j <= nn - (1 << i); ++j)
				a[i][j] = std::min(a[i - 1][j], a[i - 1][j + (1 << i - 1)]);
	}
	int query(int l, int r) {
		int k = 31 - __builtin_clz(r - l);
		return std::min(a[k][l], a[k][r - (1 << k)]);
	}
} s[2];

std::priority_queue<std::tuple<int, int, int>> q;

int query(int l, int r, bool e) {
	l = l + !e >> 1, r = r + !e >> 1;
	return s[e].query(l, r);
}
void update(int l, int r) {
	if (l == r) return;
	q.emplace(-query(l, r, l & 1), l, r);
}

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n, nn = n >> 1;
	for (int i = 0; i < n; ++i) std::cin >> p[i], ip[p[i]] = i;
	for (int i = 0; i < n; ++i) s[i & 1].a[0][i >> 1] = p[i];
	s[0].init(), s[1].init(), update(0, n);
	while (q.size()) {
		int v, l, r; std::tie(v, l, r) = q.top(); q.pop();
		int L = ip[-v], R = ip[query(L + 1, r, l & 1 ^ 1)];
		ans[tot++] = p[L], ans[tot++] = p[R];
		update(l, L), update(L + 1, R), update(R + 1, r);
	}
	for (int i = 0; i < n; ++i) std::cout << ans[i] << ' '; std::cout << '\n';
	return 0;
}